Minimally spanning iff basis
Dependencies:
A finite set $S$ minimally spans $V$ iff it is a basis.
Proof
Let $S$ be minimally spanning. Assume $S$ is linearly dependent. Therefore, $\exists v \in S, \operatorname{span}(S - \{v\}) = \operatorname{span}(S) = V$. This contradicts the fact that $S$ is minimally spanning. Therefore, $S$ is linearly independent, which makes $S$ a basis.
Let $S$ be a basis. Assume $S$ is not minimally spanning. Therefore, $\exists v \in S, \operatorname{span}(S - \{v\}) = \operatorname{span}(S) = V$. Since $S$ is linearly independent and $S - \{v\}$ spans $V$, $|S| \le |S - \{v\}| \Rightarrow |S| \le |S| - 1 \Rightarrow \bot$. Therefore, $S$ is minimally spanning.
Dependency for: None
Info:
- Depth: 6
- Number of transitive dependencies: 38
Transitive dependencies:
- /linear-algebra/vector-spaces/condition-for-subspace
- /linear-algebra/matrices/gauss-jordan-algo
- /sets-and-relations/equivalence-relation
- Group
- Ring
- Polynomial
- Integral Domain
- Comparing coefficients of a polynomial with disjoint variables
- Field
- Vector Space
- Linear independence
- Span
- Decrementing a span
- Semiring
- Matrix
- Stacking
- System of linear equations
- Product of stacked matrices
- Matrix multiplication is associative
- Reduced Row Echelon Form (RREF)
- Matrices over a field form a vector space
- Row space
- Elementary row operation
- Every elementary row operation has a unique inverse
- Row equivalence of matrices
- Row equivalent matrices have the same row space
- RREF is unique
- Identity matrix
- Inverse of a matrix
- Inverse of product
- Elementary row operation is matrix pre-multiplication
- Row equivalence matrix
- Equations with row equivalent matrices have the same solution set
- Basis of a vector space
- Linearly independent set is not bigger than a span
- Homogeneous linear equations with more variables than equations
- Rank of a homogenous system of linear equations
- Rank of a matrix